Question

You throw a rock near a cliff. you gave the rock an initial velocity vi=21.0m/s at an angle θ=62.0° above the horizontal

(a) What was the highest the rock got above from where you threw it?

(b) The rock continues its journey, misses the edge of the cliff, and lands at d=20.0m below the edge of the cliff. What was the total time taken for the rock from the moment it left your hand to where it landed below the cliff?

Answer 1

Vy = V sin θ = 21 m/s * sin 62 = 18.5 m/s

Vy - a t = 0 when it reaches its highest point

t = 18.5 / 9.80 = 1.89 s time to reach highest point

H = Vy t - 1/2 g t^2 = 21 * 1.89 - 1/2 * 9.80 * 1.89^2 = 22.2

It will reach a height of 22.2 m above the edge of the cliff

It will fall 22.2 + 20 = 42.2 m landing

S = 1/2 g t^2 total time to fall 42.2 m

t^2 = 2 * 42.2 / 9.80 = 8.61 s^2

t = 2.93 sec time to fall 42.2 m

T = 1.89 + 2.93 = 4.82 sec total time in air